Question

A new kind of tulip is produced that develops only purple or pink flowers. Assume that flower color is controlled by a single-gene locus, and that the purple allele (C) is dominant to the pink allele (c). A random sample of 1000 tulips from a large cultivated field yields 847 purple flowers, and 153 pink flowers.
a. Determine the frequency of the purple and pink alleles in this field population.
b. Estimate the proportion of all purple flowering plants that are heterozygotes and homozygotes.
a. purple allele (C) = 0.609, pink allele (c) = 0.391
b. purple homozygotes = 371, pink homozygotes = 153, heterozygotes = 476

Answer 1

Answer:

a. purple allele (C) = 0.609, pink allele (c) = 0.391

b. purple homozygotes = 371, pink homozygotes = 153, heterozygotes = 476

Explanation:

Given -

Purple flowers - 847

Pink flowers - 153

The frequency of recessive genotype i.e

$q^ 2 = \frac{153}{1000} \\q^ 2 = 0.153$

Frequency of recessive allele i.e q is equal to

$q = \sqrt{0.153} \\q = 0.391$

As per hardy Weinberg's first equilibrium equation -

$p + q = 1\\p = 1-q\\p = 1-0.391\\p = 0.609$

Frequency of purple homozygous species

$= p^2\\= 0.609^2\\= 0.371$

Number of purple homozygous species $= 0.371 * 1000= 371$

Number of pink homozygous species $= 0.153 * 1000= 153$

Heterozygous species is equal to

$(1-0.371-0.153)* 1000\\= 0.476 * 1000\\= 476$

Answer 2

Answer:

a. purple allele (C) = 0.609, pink allele (c) = 0.391

b. purple homozygotes = 371, purple heterozygotes = 476

Explanation:

According to Hardy-Weinberg equilibrium:

p + q = 1

p² + 2pq + q² = 1

where, p = frequency of dominant allele

q = frequency of recessive allele

p² = frequency of dominant homozygous genotype

2pq = frequency of heterozygous genotype

q² = frequency of homozygous recessive genotype

a. Assuming the population is in Hardy-Weinberg equilibrium,

frequency of pink flowers (cc) = q² = 153/1000 = 0.153

frequency of pink allele c = q = √0.153 = 0.391

p + q = 1 where p = frequency of dominant allele

So, frequency of purple allele C = p = 1-0.391 = 0.609

b. Purple homozygotes = CC = p² * total population =

(0.609)² * 1000 = 370.88 = 371

Purple heterozygotes = Cc = 2pq * total population =

2 * 0.609 * 0.391 * 1000 = 476.23 = 476